Sps $AB, CD$ met at $T$ and $(EAB), (ECD)$ met at $Z$. By the condition we get $\triangle ZAB \sim \triangle ZCD$, where $P$ and $Q$ reacts each other, $Z$ is the Miquel point, and $(T,P,Z,Q)$ concyclic. By angle chasing, we earn $(B,P,S,Z)$ concyclic and $\triangle ZSP \sim \triangle ZEA$ (vice versa), then since $$\angle ESR = \angle PSB = \angle PZB = \angle DZQ = \angle EZR,$$$(E,S,Z,R)$ is concyclic, as desired.

Let $(EAB)$ and $(ECD)$ meet at $X$, the spiral center of $AB\mapsto CD$. Then by the ratio condition $X$ is also the spiral center $AP\mapsto CQ$. So $X$ lies on $(CQR)$ and is thus the Miquel point of $CQDSER$, implying that $X$ lies on $(ERS)$, finishing the problem.

Essentially IMO 2005 P5